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In the footnotes of my old (numerical methods and types of errors) lecture notes there is a question to find $\sqrt{X}-\sqrt{x}$ given that $E=X-x$. I am assuming that $E$ represents the error difference. The notes are old and not very clear but it mentions that $\sqrt{X}-\sqrt{x}$ is proportional to $E/2\sqrt{X}$. Is there a way to show that?

gbd
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2 Answers2

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Since $E=X-x=(\sqrt X-\sqrt x)(\sqrt X+\sqrt x)$, when the error is small you can approximate $(\sqrt X+\sqrt x)$ by $2\sqrt X$. Hence $\sqrt X-\sqrt x$ is around $E/2\sqrt X$.

Tesla Daybreak
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$$\sqrt X-\sqrt x=\frac{X-x}{\sqrt X+\sqrt x}=\frac E{2z}$$

where

$$z:=\frac{\sqrt X+\sqrt x}2.$$

When accuracy does not matter, you can use one of $\sqrt X$ and $\sqrt x$ for $z$.